Problem: Recently Duff has been a soldier in the army. Malek is her commander.

Their country, Andarz Gu has n cities (numbered from 1 to n) and n - 1 bidirectional roads. Each road connects two different cities. There exist a unique path between any two cities.

There are also m people living in Andarz Gu (numbered from 1 to m). Each person has and ID number. ID number of i - th person is i and he/she lives in city number ci. Note that there may be more than one person in a city, also there may be no people living in the city.

<image>

Malek loves to order. That's why he asks Duff to answer to q queries. In each query, he gives her numbers v, u and a.

To answer a query:

Assume there are x people living in the cities lying on the path from city v to city u. Assume these people's IDs are p1, p2, ..., px in increasing order. 

If k = min(x, a), then Duff should tell Malek numbers k, p1, p2, ..., pk in this order. In the other words, Malek wants to know a minimums on that path (or less, if there are less than a people).

Duff is very busy at the moment, so she asked you to help her and answer the queries.

Input

The first line of input contains three integers, n, m and q (1 ≤ n, m, q ≤ 105).

The next n - 1 lines contain the roads. Each line contains two integers v and u, endpoints of a road (1 ≤ v, u ≤ n, v ≠ u).

Next line contains m integers c1, c2, ..., cm separated by spaces (1 ≤ ci ≤ n for each 1 ≤ i ≤ m).

Next q lines contain the queries. Each of them contains three integers, v, u and a (1 ≤ v, u ≤ n and 1 ≤ a ≤ 10).

Output

For each query, print numbers k, p1, p2, ..., pk separated by spaces in one line.

Examples

Input

5 4 5
1 3
1 2
1 4
4 5
2 1 4 3
4 5 6
1 5 2
5 5 10
2 3 3
5 3 1


Output

1 3
2 2 3
0
3 1 2 4
1 2

Note

Graph of Andarz Gu in the sample case is as follows (ID of people in each city are written next to them):

<image>

Solution (in CPP):

Solution:
#include <bits/stdc++.h>
using namespace std;
const int N = 1e5 + 3;
vector<int> path[N][21], adj[N];
int par[N][20], h[N];
vector<int> merge(vector<int> u, vector<int> v) {
  u.insert(u.end(), v.begin(), v.end());
  sort(u.begin(), u.end());
  if (u.size() > 10) u.resize(10);
  return u;
}
void dfs(int c, int p = 0) {
  int i, x;
  h[c] = h[p] + 1;
  par[c][0] = p;
  path[c][1] = path[p][0];
  for (i = 1; i < 20; ++i) {
    x = par[c][i - 1];
    if (par[x][i - 1] == 0) break;
    par[c][i] = par[x][i - 1];
    path[c][i + 1] = merge(path[c][i], path[x][i]);
  }
  for (i = 0; i < adj[c].size(); ++i)
    if (adj[c][i] != p) dfs(adj[c][i], c);
}
int main() {
  int a, b, n, m, i, q;
  scanf("%d%d%d", &n, &m, &q);
  for (i = 0; i < n - 1; ++i)
    scanf("%d%d", &a, &b), adj[a].push_back(b), adj[b].push_back(a);
  for (i = 1; i <= m; ++i) {
    scanf("%d", &a);
    if (path[a][0].size() < 10) path[a][0].push_back(i);
  }
  dfs(1);
  for (i = 0; i < q; ++i) {
    int u, v, k;
    scanf("%d%d%d", &u, &v, &k);
    if (h[u] > h[v]) swap(u, v);
    int diff = h[v] - h[u], i;
    vector<int> a = path[v][0];
    for (i = 0; i < 20; ++i)
      if (diff & (1 << i)) {
        a = merge(a, path[v][i + 1]);
        v = par[v][i];
      }
    if (u != v) {
      a = merge(a, path[u][0]);
      for (i = 19; i >= 0; --i) {
        if (par[u][i] != par[v][i]) {
          a = merge(path[u][i + 1], a);
          a = merge(path[v][i + 1], a);
          u = par[u][i];
          v = par[v][i];
        }
      }
      u = par[u][0];
      a = merge(a, path[u][0]);
    }
    k = min(k, (int)a.size());
    printf("%d ", k);
    for (i = 0; i < k; ++i) printf("%d ", a[i]);
    putchar('\n');
  }
}